Question

The area enclosed by the curve $x=3\cos \theta ,y=2\sin \theta ,0\le \theta \le \pi$, is (in square units)

• A. $9\pi$
• B. $6\pi$
• C. $4\pi$
• D. $3\pi$
• E. $2\pi$

Answer 1

Answer: (D)

$\frac{x}{3}=\cos \theta$
$\frac{y}{2}=\sin \theta$
$\frac{x^2}{9}+\frac{y^2}{4}=1$
$\Rightarrow$ (curve is ellipse in first and 2nd quadrant)
area $=\frac{\pi ab}{2}$
$=\frac{\pi +3\times 2}{2}=3\pi$